anonymous
  • anonymous
Identify the 31st term of an arithmetic sequence where a1 = 26 and a22 = -226. -334 -274 -284 -346
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
hint: we can write these equations: \[\Large \begin{gathered} {a_{22}} = {a_1} + 21d \hfill \\ {a_{31}} = {a_1} + 30d \hfill \\ \end{gathered} \] since the general formula is: \[\Large {a_n} = {a_1} + \left( {n - 1} \right)d\] where d is the constant of your sequence
Michele_Laino
  • Michele_Laino
using your data we can rewrite the first equation as follows: \[\Large - 226 = 26 + 21d\] please solve that equation for d
anonymous
  • anonymous
-252=21d d=-12

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anonymous
  • anonymous
is that right
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
now, substituting that value of d into the second equation, we get: \[\Large {a_{31}} = 26 + 30 \times \left( { - 12} \right) = ...?\]
anonymous
  • anonymous
-344
anonymous
  • anonymous
Thanks!
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
Given the functions f(n ) = 11 and g(n ) = -2(n - 1), combine them to create an arithmetic sequence, an, and solve for the 31st term. an = 11 - 2(n - 1); a31 = -49 an = 11 - 2(n - 1); a31 = -51 an = 11 + 2(n - 1); a31 = 71 an = 11 + 2(n - 1); a31 = 73
anonymous
  • anonymous
can u help me with this one
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
for example, let's consider the first option: we have: \[\Large {a_n} = 11 - 2\left( {n - 1} \right)\] so for n=31, we can rewrite that equationas follows: \[\Large {a_{31}} = 11 - 2 \times \left( {31 - 1} \right) = ...?\] please continue
anonymous
  • anonymous
a31= -49 right?
Michele_Laino
  • Michele_Laino
yes! that's right!
anonymous
  • anonymous
Thanks
anonymous
  • anonymous
Given an arithmetic sequence in the table below, create the explicit formula and list any restrictions to the domain. n an 1 40 2 47 3 54
anonymous
  • anonymous
can u help me with this one
anonymous
  • anonymous
Those are the options: an = 40 + 7(n - 1) where n ≥ 40 an = 40 + 7(n - 1) where n ≥ 1 an = 40 - 7(n - 1) where n ≥ 40 an = 40 - 7(n - 1) where n ≥ 1
Michele_Laino
  • Michele_Laino
Please wait: also the third option is correct, since we can write this: \[\Large {a_{31}} = 11 + 2 \times \left( {31 - 1} \right) = ...?\]
anonymous
  • anonymous
Oh
anonymous
  • anonymous
a31=11+2×(31−1)= 71
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
71 is not an answer option tho
anonymous
  • anonymous
Oh yes it is nevermind
anonymous
  • anonymous
Which one is the right one? How can I know?
Michele_Laino
  • Michele_Laino
I'm pondering...
anonymous
  • anonymous
Do I just make a guess?
Michele_Laino
  • Michele_Laino
maybe the first one, since the first option contains both f(n) and g(n), whereas the third option contains f(n) and -g(n)
anonymous
  • anonymous
Okay thanks!
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
can u help me with thise one now: Given an arithmetic sequence in the table below, create the explicit formula and list any restrictions to the domain. n an 1 40 2 47 3 54 Those are the options: an = 40 + 7(n - 1) where n ≥ 40 an = 40 + 7(n - 1) where n ≥ 1 an = 40 - 7(n - 1) where n ≥ 40 an = 40 - 7(n - 1) where n ≥ 1
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
the constant of your sequence is: \[d = 47 - 40 = 54 - 47 = ...?\]
anonymous
  • anonymous
d=7
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
:)
Michele_Laino
  • Michele_Laino
so, since the general formula, is: \[\Large {a_n} = {a_1} + \left( {n - 1} \right)d\] replace a_1 with 40 and d with 7, what do you get?
anonymous
  • anonymous
an=40+(n−1)7
anonymous
  • anonymous
whats in the n spot?
Michele_Laino
  • Michele_Laino
n is the number of terms, it is a natural number, more precisely n-1 is the number of terms of the sequence which precede a_n
anonymous
  • anonymous
Okay so is the answer the second one
Michele_Laino
  • Michele_Laino
yes! since we start to count from n=1
anonymous
  • anonymous
THANKS!
Michele_Laino
  • Michele_Laino
:)

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